+0

cos^6x=cos^4x

0
699
5

cos^6x=cos^4x

Nov 8, 2014

#2
+94619
+10

cos^6x = cos^4x     subtract cos^4x ffrom both sides

cos^6x - cos^4x =  0    faactor this

cos^4x*(cos^2x - 1) = 0

cos^4x (cosx + 1)(cosx-1)   setting each factor to 0, we have

x = 0,  pi/2,  pi , (3 pi)/2        in the interval  [0, 2pi)

More general solutions are  x = 0 + (pi)n  and   x = pi/2 + (pi/2)n    where n is an integer

Nov 8, 2014

#1
+95369
+5

$$\\cos^6x=cos^4x\\ Divide both sides by \;cos^4x\\ cos^2x=1\\ cosx=\pm1 At this point I just think about the unit circle\\ cos of an angle is given by the x value (not the angle x)\\ The x value is 1 at 0 and it is -1 at 180 degrees\\ So \\ x=180n degrees where  n\in Z\\ or\\ x=n\pi radians where  n\in Z\\$$

.
Nov 8, 2014
#2
+94619
+10

cos^6x = cos^4x     subtract cos^4x ffrom both sides

cos^6x - cos^4x =  0    faactor this

cos^4x*(cos^2x - 1) = 0

cos^4x (cosx + 1)(cosx-1)   setting each factor to 0, we have

x = 0,  pi/2,  pi , (3 pi)/2        in the interval  [0, 2pi)

More general solutions are  x = 0 + (pi)n  and   x = pi/2 + (pi/2)n    where n is an integer

CPhill Nov 8, 2014
#3
+95369
+5

Okay I missed some answers - thanks for picking that up chris but your presentation looks strange.

isn't

x = 0 + (pi)n  and   x = pi/2 + (pi/2)n

really just

$$x=\frac{n\pi}{2}\qquad where \qquad n\in Z$$

???

Nov 8, 2014
#4
+94619
+5

Yeah..... your presenttion is better....we have to be careful in these trig equations about "dividing away" solutions....it's analagous to "throwing away" roots in a polynomial.....

Nov 8, 2014
#5
+95369
0

YES WE DO  !!!    ME MORE THAN YOU !!!      LOL       :)))))

Nov 8, 2014